In the pictures below, you can see several
types of Dices. It is very easy to assume that they are used in many different
types of games. In this problem, the face of the dices will be like those on the
first figure, that means each will have some dots it each face, which will
indicate the value of that particular face. But the looks of the dices will be
like those on the second figure, that means each dice may have 2, 3, 4, 5 up to 50 different faces. A dice with p faces have face values 1, 2, 3 … p. When a dice is thrown all
the faces are equally likely to come on top. For example an ordinary dice in
figure1 has face values 1, 2, 3, 4, 5,
6 and the probability that face with value 1 will be on top is 1/6. The same is the probability for
the other faces coming on top.


Figure 1: Two conventional dices  Figure 2: Dices of different shape 
The situation becomes complicated when you
throw a dice more than once dice. For example when you throw a dice twice, for
any arbitrary incident the summation of the top two faces can be within the
range (inclusive) 2...12. But their
probability is not equally likely or 1/12. The probability that the sum will
be two is 1/36, but the probability
that some will be three is 2/36 and
so on. Given the description of a dice throw, that’s how many faces it have, how
many throws will be made and value for the sum, you will have to find the
probability of obtaining that sum.
The input will contain several lines of input. Each line contains three integers F (The number of faces of the dice thrown, 1<=F<=50), N (The number of throws and 0<=N<=50) and S (The summation value for which you to find out the probability, 0<S<=4000).
For each line of input you will have to produce one line of output which will be of the form a/b (Here b = F^{N}). Here a/b means the probability value. Note that we don’t need the floating point value of probability.
(World Final Warmup Contest, Problem setter: Shahriar
Manzoor)
“He He!!! Why does
this guy like probability so much? :)”